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4 years ago

10

Find the x- and y-intercepts of 3x + 6y = 18

1 answer:

djyliett [7]4 years ago

3 0

In order to find the x and y intercepts of an equation, you can plug in 0 for the opposite variable. Let us solve for the intercepts now.

<span>x-intercept

</span>We are going to plug in 0 for y and the solve for x:

$3x+6y=18$
$3x+6(0)=18$
$3x=18$
$x=6$

So now we know the x-intercept of the line is: (6, 0)
<span>
y-intercept</span>

We are going to plug in 0 for x and solve for y:

$3x+6y=18$
$3(0)+6y=18$
$6y=18$
$y=3$

So now we know that the y-intercept for the line is: (0, 3)

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Solve it :<br> 4xy2+5x+6xy2−2y2<br><br> pls finish this fast

photoshop1234 [79]

Answer:

20xy + 5x -4y

Step-by-step explanation:

I hope this helps you

6 0

3 years ago

Caleb got $60 for his birthday and put it in his savings account. After he deposited the money, he has $139. How much money was

seropon [69]

A. 60-x=139
B. 139-60=79
79=x

3 0

3 years ago

Kay [80]

Answer:

$\frac{(82)(2.4)^2}{104.139} \leq \sigma^2 \leq \frac{(82)(2.4)^2}{62.132}$

$4.535 \leq \sigma^2 \leq 7.602$

Now we just take square root on both sides of the interval and we got:

$2.2 \leq \sigma \leq 2.8$

And the best option would be:

A. 2.2 < σ < 2.8

Step-by-step explanation:

Information provided

$\bar X=32.1$ represent the sample mean

$\mu$ population mean

s=2.4 represent the sample standard deviation

n=83 represent the sample size

Confidence interval

The confidence interval for the population variance is given by the following formula:

$\frac{(n-1)s^2}{\chi^2_{\alpha/2}} \leq \sigma^2 \leq \frac{(n-1)s^2}{\chi^2_{1-\alpha/2}}$

The degrees of freedom are given by:

$df=n-1=83-1=82$

The Confidence is given by 0.90 or 90%, the value of $\alpha=0.1$ and $\alpha/2 =0.05$ , the critical values for this case are:

$\chi^2_{\alpha/2}=62.132$

$\chi^2_{1- \alpha/2}=104.139$

And replacing into the formula for the interval we got:

$\frac{(82)(2.4)^2}{104.139} \leq \sigma^2 \leq \frac{(82)(2.4)^2}{62.132}$

$4.535 \leq \sigma^2 \leq 7.602$

Now we just take square root on both sides of the interval and we got:

$2.2 \leq \sigma \leq 2.8$

And the best option would be:

A. 2.2 < σ < 2.8

3 0

3 years ago

What is the square root of -1?

Natali [406]

If you work with real numbers then the square root of - 1 CANNOT be defined.

If you work with complex numbers then the square root of - 1 is equal to the imaginary unit(which is symbolized as i)

6 0

3 years ago

Read 2 more answers

Which expression is equivalent?

chubhunter [2.5K]

It'll be 8 |a^3|.

we don't use modulus on 8 cuz it is obviously positive but we do not know about a^6 when rooted it can be +ve or -ve.

6 0

3 years ago